A 1000.1 g aluminium rod at 83.1 °C is placed in 1224.8 ml of water at 2.8 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.

Question

A 1000.1 g aluminium rod at 83.1 °C is placed in 1224.8 ml of water at 2.8 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.

BlackTom AI · Accepted Answer

Heat lost by the aluminium equals heat gained by the water (conservation of energy), so m_Al c_Al (T_Al,i − T_f) = m_w c_w (T_f − T_w,i).

1) Relevant knowledge points and concepts
- Conservation of energy (no heat lost to surroundings): heat lost by hot body = heat gained by cold body.
- Specific heat relation: Q = m c ΔT.
- Convert masses/volumes to kilograms: 1000.1 g = 1.0001 kg; 1224.8 mL water ≈ 1.2248 kg.
- Plug given specific heats: c_Al = 880 J kg−1 K−1, c_w = 4186 J kg−1 K−1.

2) Complete reasoning (algebra)
m_Al c_Al (T_Al,i − T_f) = m_w c_w (T_f − T_w,i)
Rearrange to solve for T_f:
(m_Al c_Al + m_w c_w) T_f = m_Al c_Al T_Al,i + m_w c_w T_w,i
T_f = (m_Al c_Al T_Al,i + m_w c_w T_w,i) / (m_Al c_Al + m_w c_w)

3) Calculations with intermediate steps
m_Al c_Al = 1.0001 kg × 880 J kg−1 K−1 = 880.088 J K−1
m_w c_w = 1.2248 kg × 4186 J kg−1 K−1 = 5127.0128 J K−1
Numerator:
m_Al c_Al T_Al,i = 880.088 × 83.1 = 73,135.3128 J
m_w c_w T_w,i = 5127.0128 × 2.8 = 14,355.63584 J
Sum (numerator) = 73,135.3128 + 14,355.63584 = 87,490.94864 J
Denominator = 880.088 + 5127.0128 = 6007.1008 J K−1
T_f = 87,490.94864 / 6007.1008 = 14.564588… °C

4) Why the correct answer is correct
The result follows directly from energy conservation and the definition of specific heat. The final temperature lies between the initial temperatures (2.8 °C and 83.1 °C) and is much closer to the cold water temperature because the water has a much larger heat capacity (m c) than the aluminium.

5) Other options
Not applicable (numerical question with one correct result). Any alternative numerical value would require different input values or neglect of energy conservation.

6) Conclusion
Rounding the computed final temperature to two decimal places gives T_f = 14.56 °C, which confirms the provided answer 14.56.

A 1000.1 g aluminium rod at 83.1 °C is placed in 1224.8 ml of water at 2.8 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.数值题

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